What is the tangent of f(x)=x^3,\at x=-3 ?

The tangent of f(x)=x^3,\at x=-3 is y=27x+54

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tangent of f(x)=x^3,\at x=-3

tangent of

f (x) = x^{3}, at x = - 3

y = 27 x + 54

Solution steps

Find the tangent point:

(- 3, - 27)

Find the slope of

f (x) = x^{3} : \frac{df ( x )}{d x} = 3 x^{2}

Compute the slope of

f (x) = x^{3}

x = - 3 : m = 27

Find the line with slope m=

27

and passing through

(- 3, - 27) : y = 27 x + 54

y = 27 x + 54

x \to - 2 lim (x - 2)

\frac{d}{d x} (2 + x)

\int e^{- \frac{1}{2}} d x

\frac{d y}{d x}, y^{5} = x^{6}

2 y^{^{'}} + y = 3 t^{2}

What is the tangent of f(x)=x^3,\at x=-3 ?
The tangent of f(x)=x^3,\at x=-3 is y=27x+54